Question

A 1-kW radio transmitter operates at a frequency of 880 Hz. How many photons per second does it emit?
A. $1.71\times {{10}^{21}}$
B. $1.71\times {{10}^{30}}$
C. $6.02\times {{10}^{23}}$
D. $2.85\times {{10}^{26}}$

Answer 1

The relation between number of photons emitted and energy of the photon and it is as follows.
$\text{Number of photons released=}\frac{\text{1}}{\text{energy of the photon}}$
E = h$\mu$
Here E = energy of the radio transmitter
H = Planck’s constant = $6.02\times {{10}^{23}}$ Js
$\mu$ = frequency of the radio transmitter

Complete answer:
- In the question it is given to calculate how many photons per second a 1-kW radio transmitter operates.
- First we have to calculate the energy of the radio transmitter.
- The formula to calculate the energy of the radio transmitter as follows.
E = h$\mu$
Here E = energy of the radio transmitter
H = Planck’s constant = $6.02\times {{10}^{23}}$ Js
$\mu$ = frequency of the radio transmitter = 880 Hz
- Substitute the known values in the above formula to calculate the energy of the radio transmitter.
E = ($6.02\times {{10}^{23}}$ ) ( 880) = 5.831$\times {{10}^{-31}}$ J
- Now by using the below formula we can calculate the number of photons released per second.

& \text{Number of photons released=}\frac{\text{1}}{\text{energy of the photon}} \\\ & =\frac{1}{5.831\times {{10}^{-31}}} \\\ & =1.71\times {{10}^{30}} \\\ \end{aligned}$$ \- Therefore the number of photons emitted per second by a 1-kW radio transmitter of 880 Hz is$1.71\times {{10}^{30}}$ . **So, the correct option is B.** **Note:** Energy of the photon is inversely proportional to the number of photons emitted by it. Radio transmitter is an electronic device that releases electronic magnetic rays in the form of a signal to other electronic devices.